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7 February, 01:27

Suppose the number of radios in a household has a binomial distribution with parameters n=11 and p=40%. Find the probability of a household having: a. 1 or 9 radios b. 7 or fewer radios c. 5 or more radios d. fewer than 9 radios e. more than 7 radios

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  1. 7 February, 01:42
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    Step-by-step explanation:

    The formula for binomial distribution is expressed as

    P (x=r) = nCr * q^ (n-r) * p^r

    From the information given,

    n = 11

    p = 40% = 40/100 = 0.4

    q = 1 - 0.4 = 0.6

    x represent the number of radios

    a) P (x = 1) or P (x = 9)

    P (x = 1) = 11C1 * 0.6^ (11-1) * 0.4^1

    P (x = 1) = 0.027

    P (x = 9) = 11C9 * 0.6^ (11-9) * 0.4^9

    P (x = 9) = 0.0052

    P (x = 1) or P (x = 9) = 0.027 + 0.0052 = 0.0322

    b) P (x lesser than or equal to 7) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) + P (x = 6)

    P (x = 0) = 11C0 * 0.6^ (11-0) * 0.4^0 = 0.004

    P (x = 1) = 0.027

    P (x = 2) = 11C2 * 0.6^ (11-2) * 0.4^2 = 0.089

    P (x = 3) = 11C3 * 0.6^ (11-3) * 0.4^3 = 0.177

    P (x = 4) = 11C4 * 0.6^ (11-4) * 0.4^4 = 0.24

    P (x = 5) = 11C5 * 0.6^ (11-5) * 0.4^5 = 0.22

    P (x = 6) = 11C6 * 0.6^ (11-6) * 0.4^6 = 0.15

    P (x = 7) = 11C7 * 0.6^ (11-7) * 0.4^7 = 0.15

    P (x lesser than or equal to 7) = 0.004 + 0.027 + 0.089 + 0.177 + 0.24 + 0.22 + 0.15 + 0.07 = 0.977

    c) P (x greater than or equal to 5) = 1 - P (x lesser than or equal to 4) = 1 - (0.004 + 0.027 + 0.089 + 0.177 + 0.24) = 1 - 0.537 = 0.463

    d) P (x lesser than 9) = P (x lesser than or equal to 7) + P (x = 8)

    P (x = 8) = 11C8 * 0.6^ (11-8) * 0.4^8 = 0.02

    P (x lesser than 9) = 0.977 + 0.02 = 0.997

    e. P (x greater than 7) = 1 - P (x lesser than or equal to 7) = 1 - 0.977 = 0.023
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